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Re: IDL hilbert() function [message #63175]

Mon, 03 November 2008 01:16

Wout De Nolf
Messages: 194
Registered: October 2008

Senior Member

On Sat, 1 Nov 2008 02:14:54 -0700 (PDT), lecacheux.alain@wanadoo.fr
wrote:

> Is this function actually computing the Hilbert transform ?
> The Hilbert transform is known to be idempotent, i.e. H(H(x)) = -x.
> However, by applying the IDL function, one get for instance :
> IDL> print, hilbert (hilbert (indgen(8)))
> ( 6.00000, 0.000000)( 7.00000, 0.000000)
> ( 4.00000, 0.000000)( 5.00000, 0.000000)
> ( 2.00000, 0.000000)( 3.00000, 0.000000)
> ( 0.000000, 0.000000)( 1.00000, 0.000000)

It works for this:
x=findgen(180)/90.*!pi
plot,x,hilbert(hilbert(sin(x))),/xs
oplot,x,-sin(x),psym=1

Not for this (flips):
plot,hilbert(hilbert((indgen(1000))))
oplot,indgen(1000),psym=1

I'm not sure why, but check the hilbert.pro in the IDL-lib directory
to see how it's implemented.

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		Re: IDL hilbert() function By: Wout De Nolf on Mon, 03 November 2008 01:16
		Re: IDL hilbert() function By: ed.schmahl on Mon, 03 November 2008 10:26

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